Question: Divide the following complex numbers. $ \dfrac{-18-13i}{-4-i}$
Answer: We can divide complex numbers by multiplying both numerator and denominator by the denominator's complex conjugate , which is ${-4+i}$ $ \dfrac{-18-13i}{-4-i} = \dfrac{-18-13i}{-4-i} \cdot \dfrac{{-4+i}}{{-4+i}} $ We can simplify the denominator using the fact $(a + b) \cdot (a - b) = a^2 - b^2$ $ \dfrac{(-18-13i) \cdot (-4+i)} {(-4-i) \cdot (-4+i)} = \dfrac{(-18-13i) \cdot (-4+i)} {(-4)^2 - (-1i)^2} $ Evaluate the squares in the denominator and subtract them. $ \dfrac{(-18-13i) \cdot (-4+i)} {(-4)^2 - (-1i)^2} = $ $ \dfrac{(-18-13i) \cdot (-4+i)} {16 + 1} = $ $ \dfrac{(-18-13i) \cdot (-4+i)} {17} $ Note that the denominator now doesn't contain any imaginary unit multiples, so it is a real number, simplifying the problem to complex number multiplication. Now, we can multiply out the two factors in the numerator. $ \dfrac{({-18-13i}) \cdot ({-4+i})} {17} = $ $ \dfrac{{-18} \cdot {(-4)} + {-13} \cdot {(-4) i} + {-18} \cdot {1 i} + {-13} \cdot {1 i^2}} {17} $ Evaluate each product of two numbers. $ \dfrac{72 + 52i - 18i - 13 i^2} {17} $ Finally, simplify the fraction. $ \dfrac{72 + 52i - 18i + 13} {17} = \dfrac{85 + 34i} {17} = 5+2i $